GATE CS 2017 Set 1 — Question 36

MCQ+2 / -0.67MediumFunctions & RecursionC ProgrammingProgramming & Data StructuresData Types & Operators

Programming & Data Structures → C Programming → Data Types & Operators

Last updated 5 September 2026

Question

Consider the C functions foo and bar given below: int foo(int val) { int x = 0; while(val > 0) { x = x + foo(val--); } return val; } int bar(int val) { int x = 0; while(val > 0) { x = x + bar(val-1); } return val; } Invocations of foo(3) and bar(3) will result in :

Correct answer (C) Abnormal termination and infinite loop respectively. Solution Analysis of foo(3) : Inside foo(3), the loop condition is val > 0 (3 > 0). The body executes x = x + foo(val--); . The argument to the recursive call is val-- . The post-decrement operator returns the current value of val (which is 3) and then decrements val to 2. So, foo(3) calls foo(3) recursively. This creates an infinite recursion: foo(3) -> foo(3) -> foo(3) ... Infinite recursion leads to a Stack Overflow, which causes Abnormal termination . Analysis of bar(3) : Inside bar(3), the loop condition is val > 0 (3 > 0). The body executes x = x + bar(val-1); . bar(3) calls bar(2) . bar(2) calls bar(1) . bar(1) calls bar(0) . bar(0) checks val > 0 (0 > 0), which is false, and returns 0. Control returns to bar(1) . x becomes x + 0 . The loop checks val > 0 (1 > 0) again. Note that val is never modified inside the bar function (it is passed by value, and val-1 does not change the local val). So bar(1) stays in the while(1 > 0) loop forever, repeatedly calling bar(0) and adding 0 to x . This is an Infinite loop . Thus, foo(3) results in abnormal termination, and bar(3) results in an infinite loop. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (C) Abnormal termination and infinite loop respectively. Solution Analysis of foo(3) : Inside foo(3), the loop condition is val > 0 (3 > 0). The body executes x = x + foo(val--); . The argument to the recursive call is val-- . The post-decrement operator returns the current value of val (which is 3) and then decrements val to 2. So, foo(3) calls foo(3) recursively. This creates an infinite recursion: foo(3) -> foo(3) -> foo(3) ... Infinite recursion leads to a Stack Overflow, which causes Abnormal termination . Analysis of bar(3) : Inside bar(3), the loop condition is val > 0 (3 > 0). The body executes x = x + bar(val-1); . bar(3) calls bar(2) . bar(2) calls bar(1) . bar(1) calls bar(0) . bar(0) checks val > 0 (0 > 0), which is false, and returns 0. Control returns to bar(1) . x becomes x + 0 . The loop checks val > 0 (1 > 0) again. Note that val is never modified inside the bar function (it is passed by value, and val-1 does not change the local val). So bar(1) stays in the while(1 > 0) loop forever, repeatedly calling bar(0) and adding 0 to x . This is an Infinite loop . Thus, foo(3) results in abnormal termination, and bar(3) results in an infinite loop.

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